4 The Schr¨odinger wave equation - UVicjwillis/teaching/phys215/phys215_lecture4.pdf ·...

4 THE SCHRODINGER WAVE EQUATION 1

4 The Schrodinger wave equation

We have noted in previous lectures that all particles, both light and matter, can be described as alocalised wave packet.

• De Broglie suggested a relationship between the effective wavelength of the wave functionassociated with a given matter or light particle its the momentum. This relationship wassubsequently confirmed experimentally for electrons.

• Consideration of the two slit experiment has provided an understanding of what we can andcannot achieve with the wave function representing the particle: The wave function Ψ is notobservable. According to the statistical interpretation of Born, the quantity Ψ∗Ψ = |Ψ2| isobservable and represents the probability density of locating the particle in a given elementalvolume.

To understand the wave function further, we require a wave equation from which we can studythe evolution of wave functions as a function of position and time, in general within a potentialfield (e.g. the potential fields associated with the Coulomb or strong nuclear force).

As we shall see, manipulation of the wave equation will permit us to calculate “most probable”values of a particle’s position, momentum, energy, etc. These quantities form the study of me-chanics within classical physics. Our quantum theory has now become quantum mechanics –the description of mechanical physics on the quantum scale. The particular sub-branch of quantummechanics accessible via wave theory is sometimes referred to as wave mechanics.

The time–dependent Schrodinger wave equation is the quantum wave equation

ih∂Ψ(x, t)

∂t= − h2

2m

∂2Ψ(x, t)

∂x2+ V (x, t) Ψ(x, t), (1)

where i =√−1, m is the mass of the particle, h = h/2π, Ψ(x, t) is the wave function representing the

particle and V (x, t) is a potential energy function. It is straight forward to extend the Schrodingerequation to three spatial dimensions

ih∂Ψ

∂t= − h2

2m

(∂2Ψ

∂x2+∂2Ψ

∂y2+∂2Ψ

∂z2

)+ V Ψ. (2)

The Schrodinger equation is postulated – it is not derived – yet the predictions based upon it areverified via experiment. It provides a correct description of physical observables. The same couldbe said for Newton’s equation

F = md2x

dt2. (3)


This is another example of an equation that is postulated – not derived – yet provides a correctdescription of physical observables. However, Newton’s equation is valid at macroscopic scales,whereas the Schrodinger is valid at microscopic, or quantum scales.

Note that the Schrodinger equation as presented here is non–relativistic. A relativistic form ofthe Schrodinger equation exists but is not considered in this course.

4.1 Investigating the Schrodinger equation

The Schrodinger is nominally similar to the classical wave equation

∂2Ψ

∂x2=

1

v2

∂2Ψ

∂t2, (4)

with the exception that the Schrodinger equation contains only a 1st order time derivative. This isthe first clue that wave solutions to the Schrodinger equation will not be identical to solutions ofthe classical wave equation.

T–Rex Example 6.1: Is the Schrodinger equation linear?

If the Schrodinger equation is linear, then, if Ψ1 and Ψ2 represent valid solutions, there must exista valid solution of the form

Ψ(x, t) = aΨ1(x, t) + bΨ2(x, t), (5)

where a and b are (real or complex) constants. The derivatives of Ψ may be written as

∂Ψ

∂t= a

∂Ψ1

∂t+ b

∂Ψ2

∂t

∂Ψ

∂x= a

∂Ψ1

∂x+ b

∂Ψ2

∂x

∂2Ψ

∂x2= a

∂2Ψ1

∂x2+ b

∂2Ψ2

∂x2. (6)

Substituting these expressions into the Schrodinger equation for Ψ, we obtain

ih

(a∂Ψ1

∂t+ b

∂Ψ2

∂t

)= − h2

2m

(a∂2Ψ1

∂x2+ b

∂2Ψ2

∂x2

)+ V (aΨ1 + bΨ2). (7)

Rearranging this expression yields

a

(ih∂Ψ1

∂t+h2

2m

∂2Ψ1

∂x2− V Ψ1

)= b

(ih∂Ψ2

∂t+h2

2m

∂2Ψ2

∂x2− V Ψ2

). (8)


As Ψ1 and Ψ2 are valid solutions to the Schrodinger equation, then the terms within the parenthesesare exactly zero, the equation makes sense and Ψ must also be a valid solution to the wave equation.Given the Schrodinger equation is linear we know that the principle of superposition is valid andwe can confidently create wave packets from linear wave superposition that are themselves validsolutions of the Schrodinger equation. Recall from Lecture 3 that wave superposition, demonstratedas interference fringes, is required to describe the results of the two-slit experiment.

T–Rex Example 6.2: Investigating valid solutions to the Schrodinger equation

In Lecture 3 we investigated solutions to the classical wave equation of the form

Ψ(x, t) = A sin(kx− ωt+ φ) (9)

where k is the wave number, ω is the angular frequency and φ is a phase constant. The waveis moving in the positive x–direction. This “classical” wave is not a valid solution to the time–dependent Schrodinger equation.

We start by considering the three derivative terms

∂Ψ

∂t= −ωA cos(kx− ωt)

∂Ψ

∂x= kA cos(kx− ωt)

∂2Ψ

∂x2= −k2A sin(kx− ωt) = −k2Ψ. (10)

Inserting these relations into the time–dependent Schrodinger equation we obtain

−ihω cos(kx− ωt) =

(h2k2

2m+ V

)Ψ

=

(h2k2

2m+ V

)A sin(kx− ωt), (11)

which is never satisfied (cos x 6= sinx).

We now consider a more general wave function

Ψ(x, t) = Aei(kx−ωt) = A[cos(kx− ωt) + i sin(kx− ωt)]. (12)

We note that this general wave solution is complex yet still represents a wave moving in the positivex–direction. In general, the amplitude A can also be complex. In order to demonstrate that this


wave function is a valid solution to the time–dependent Schrodinger equation, we follow the sameapproach as before, i.e.

∂Ψ

∂t= −iωAei(kx−ωt) = −iωΨ

∂Ψ

∂x= ikΨ

∂2Ψ

∂x2= i2k2Ψ = −k2Ψ. (13)

Inserting these relations into the time–dependent Schrodinger equation we obtain

ih(−iωΨ) = − h2

2m(−k2Ψ) + VΨ

(hω − h2k2

2m+ V

)Ψ = 0. (14)

We can understand this equation by noting that E = hf = hω and p = hk to obtain(E − p2

2m− V

)Ψ = 0. (15)

The term in parentheses represents the conservation of energy (i.e. E−KE−V = 0) and thereforeAei(kx−ωt) represents a valid solution to the time–dependent Schrodinger equation.

4.2 Probability and normalisation

In the previous lecture we introduced the idea of probabilities based upon the wave function. Theprobability P (x) dx of observing a particle between x and x+ dx is

P (x) dx = Ψ∗(x, t)Ψ(x, t) dx. (16)

The probability of observing the particle between x1 and x2 is

P =∫ x2

x1

Ψ∗Ψ dx. (17)

The particle must exist somewhere; this condition is imposed by normalising the wave functionaccording to ∫ ∞

−∞Ψ∗(x, t)Ψ(x, t) = 1. (18)


T–Rex Example 6.4 Normalising the wave function

Consider the wave function Ψ(x, t) = Ae−α|x|. Normalise the wave function and determine theprobability of observing the particle between 0 < x < 1/α, and between 1/α < x < 2/α.

The wave function is normalised as follows∫ ∞

−∞A2e−2α|x| dx = 1. (19)

As the wave function is symmetric about zero we may re–write it as

2∫ ∞

0A2e−2α|x| dx = 1 =

2A2

−2α[e−2αx]∞0

1 =−A2

α(0− 1) =

A2

α. (20)

The coefficient A =√α and the wave function may be written as

Ψ =√αe−α|x|. (21)

The probability of locating the particle between 0 < x < 1/α may be written as

P =∫ 1/α

0αe−2αx

=α

−2α[e−2α]

1/α0

= −1

2(e−2 − 1) = 0.432. (22)

Compute the value of the probability of localising the particle between 1/α < x < 2/α as anexercise.

4.3 Properties of valid wave functions

In order that the wave function corresponds to physical (i.e. real) situations, it must satisfy anumber of extra conditions in addition to the Schrodinger equation - these are sometimes referredto as boundary conditions:

1. In order to avoid infinite probabilties, Ψ must be finite everywhere.

2. In order to avoid multiple probability values, Ψ must take a single value at each position andtime.

3. For finite potentials, ∂Ψ/∂x must also be continuous (things are different when V is infinite).

4. In order to normalise the wave function one must have limx→±∞ Ψ → 0.


4.3.1 The free wave function

The wave function Ψ(x, t) = Aei(kx−ωt) represents a valid solution to the Schrodinger equation. Thewave function is referred to as the free wave function as it represents a particle experiencing zeronet force (constant V ). However, the free wave function cannot be normalised as it is a continuous(not localised) wave, i.e. ∫ ∞

−∞Ψ∗(x, t)Ψ(x, t) =

∫ ∞

−∞A2dx

= ∞. (23)

One can understand this result via the uncertainty principle: the free wave function possesses adefinite value k and ω (∆k = ∆ω = 0). Thereforewe must have ∆x = ∆t = ∞. Despite thiscondition, the free wave function remains a useful basic solution to the Schrodinger equation as allvalid wave functions can be constructed from a sum or integral of free wave functions, for example

ΨTOT =∫ ∞

−∞e−x2/2σei(kx−ωt) dx, (24)

represents a Gaussian wave packet.

4.4 The time independent Schrodinger equation

We have so far considered the time dependent Schrodinger equation. However, in the case wherethe potential term is constant, i.e. V (x, t) = V (x), the Schrodinger equation can be separated intoposition and time dependent components

Ψ(x, t) = ψ(x)f(t). (25)

Inserting this wave function into the time dependent Schrodinger equation, we obtain

ihψ(x)∂f(t)

∂t= − h

2f(t)

2m

∂2ψ(x)

∂x2+ V (x)ψ(x)f(t). (26)

Dividing by ψ(x)f(t) yields

ih1

f(t)

df(t)

dt= − h2

2m

1

ψ(x)

d2ψ(x)

dx2+ V (x). (27)

The lhs of this equation is a function of time only, the rhs is a function of position only; thereforeeach must be equal to a constant value, named B. To determine the value of B, we consider the


lhs only

ih1

f(t)

df(t)

dt= B

ih∫ df

f=

∫B dt

ih ln f = Bt+ C, (28)

where C is an integration constant which can be set to zero by specifying f(t = 0) = 0. Therefore

ln f =Bt

ih

f(t) = eBt/ih = e−iBt/h. (29)

If we compare this result to the free particle wave function we note that f(t) = e−iωt and thatB = hω = E, the total energy. Note that this can also be seen from a dimensional analysis of theterm f(t) = e−iBt/h; t has units of seconds and h has units of energy times time – therefore B musthave units of energy. This is a general result. We may now write

ih1

f(t)

df(t)

dt= E, (30)

and we can now form the time independent Schrodinger equation as

− h2

2m

d2ψ(x)

dx2+ V (x)ψ(x) = E ψ(x). (31)

Returning to the general form of the wave function we may write

Ψ(x, t) = ψ(x)e−iωt (32)

The probability density of this wave function is

Ψ∗Ψ = ψ2(x)(eiωte−iωt)

= ψ2(x). (33)

The probability distribution is constant in time. In classical physics this phenomena is referred toas a standing wave. In quantum mechanics, we refer to this as a stationary state. Therefore wecan use the time independent Schrodinger equation to study stationary or stable states in quantumsystems.


4.5 Expectation values

If the wave function represents localised particles, we require a method for computing the predictedvalues of observable quantities (e.g. position, momentum, total energy) from the wave function.

The determination of observable quantities from the wave function is based upon expectationvalues. The idea of an expectation value is based upon the statistical interpretation of the wavefunction. If we make many measurements of a particle described by a given wave function, theaverage of many measurements will converge upon the expectation value.

Any observable quantity for which we can compute an expectation value from the wave functionis referred to as a physical observable. Physical observables must be represented by real – notimaginary – numbers.

Imagine that we observe the position of a particle constrained to move along the x–axis. We observethe particle at x1, N1 times, at x2, N2 times, and so on. The average position of the particle is

x =x1N1 + x2N2 + x3N3 + x4N4 + · · ·

N1 +N2 +N3 +N4 + · · ·=

∑i xiNi∑iNi

. (34)

If P (x) describes the probability of observing the particle at some x (distributed continuously) wemay re–write the above equation in integral form

x =

∫∞−∞ xP (x) dx∫∞−∞ P (x) dx

. (35)

Within our quantum mechanical formalism P (x) = Ψ∗(x, t)Ψ(x, t) and the expectation value of xis now

〈x〉 =

∫∞−∞ Ψ∗(x, t)xΨ(x, t)dx∫∞−∞ Ψ∗(x, t)Ψ(x, t)dx

=∫ ∞

−∞Ψ∗(x, t)xΨ(x, t)dx, (36)

for a normalised wave function. the expectation value of a general function g(x) for a normalisedwave function is

〈g(x)〉 =∫ ∞

−∞Ψ∗(x, t) g(x) Ψ(x, t)dx. (37)

The expectation value of a stationary state may be written as

〈x〉 =∫ ∞

−∞ψ∗(x)eiωt xψ(x)e−iωtdx

=∫ ∞

−∞ψ∗(x)xψ(x)dx. (38)

When we state that the wave function provides a complete description of a physical system, wemean that the expectation value of physical observables (real quantities) can be computed usingthe wave function. The wave function cannot provide the value of individual measurements.


4.5.1 The momentum operator

How do we compute the expectation value of the momentum? Any knowledge of the momentummust also be consistent with the uncertainty principle. Consider the free particle wave function,Ψ(x, t) = ei(kx−ωt). Taking the derivative w.r.t. x, we have

∂Ψ

∂x=

∂

∂x[ei(kx−ωt)] = ikei(kx−ωt) = ikΨ. (39)

The wave number k = p/h, such that∂Ψ

∂x= i

p

hΨ, (40)

and rearranging the equation yields

p[Ψ(x, t)] = −ih ∂∂x

[Ψ(x, t)]. (41)

We therefore define the momentum operator as

p = −ih ∂∂x. (42)

In mathematical language, an operator transforms one function into another, e.g. the operator Aoperates on f(x) such that A f(x) = g(x). Every physical observable has an associated operatorthat is used to determine the expectation value of the observable. The general formalism is

〈A〉 =∫ ∞

−∞Ψ∗(x, t) AΨ(x, t) dx. (43)

The expectation value of the momentum is therefore

〈p〉 = −ih∫ ∞

−∞Ψ∗(x, t)

∂Ψ(x, t)

∂xdx. (44)

As we have seen earlier, position x is its own operator.

4.5.2 The energy operator

We next consider the total energy of the wave function. Take the time derivative of the free particlewave function

∂Ψ

∂t=

∂

∂tei(kx−ωt) = −iωei(kx−ωt) = −iωΨ. (45)

Substituing ω = E/h we have

E[Ψ(x, t)] = ih∂

∂t[Ψ(x, t)], (46)


and we define the energy operator as

E = ih∂

∂t. (47)

The expectation value of the energy is then

〈E〉 = ih∫ ∞

−∞Ψ∗(x, t)

∂Ψ(x, t)

∂tdx. (48)

Though the above results have been generated using the free particle wave function, they are validin general as well.

T–Rex example 6.6: Use energy conservation to generate the Schrodinger equation

The total energy of a particle may be written as

E = K + V =p2

2m+ V. (49)

We now permit the operators associated with each observable to act upon the wave function.Considering the lhs of the above equation

EΨ = ih∂Ψ

∂t. (50)

The rhs of the equation becomes

[1

2m(p)2 + V

]Ψ =

1

2m

(−ih ∂

∂x

)2

Ψ + V Ψ

= − h2

2m

∂2Ψ

∂x2+ V Ψ. (51)

Equating the lhs with the rhs of the original energy conservation equation we obtain the timedependent Schrodinger equation.

ih∂Ψ

∂t= − h2

2m

∂2Ψ

∂x2+ V Ψ. (52)

It is important to note that this is not a derivation of the Schrodinger equation. Instead it showsthe consistency of the energy and momentum terms just defined.

4.6 The infinite square well potential

We now wish to solve the time independent Schrodinger equation for several simple potentials. Indoing so we will investigate the behaviour of the quantum wave function and the expected value of


physical observables. The infinite square well potential is essentially a particle in a box. However,on this occasion our analysis will be more rigourous.

The infinite square well describes the potential term

V (x) = ∞ x ≤ 0, x ≥ L

= 0 0 < x < L, (53)

and the particle is restricted to the interval 0 < x < L. This can be seen clearly by considering thetime independent Schrodinger equation

− h2

2m

d2ψ(x)

dx2+ V (x)ψ(x) = E ψ(x). (54)

When V (x) = ∞ we must have ψ(x) = 0 (to avoid an infinity in the equation) and thereforeψ∗ψ = 0 with the consequence that the particle cannot exist in this region. When V (x) = 0, ψ(x)now represents the free particle wave function and we re–write the time independent Schrodingerequation as

− h2

2m

d2

dx2ψ(x) = Eψ(x)

d2ψ(x)

dx2= −2mE

h2 ψ(x). (55)

We note that p = hk and E = p2/2m. Therefore, p2 = 2mE and k2 = (2mE)/h2 and we mayre–write the above equation as

d2ψ(x)

dx2= −k2ψ(x). (56)

A general solution to this equations takes the form

ψ(x) = A sin kx+B cos kx. (57)

However, does such a solution match the boundary conditions for an acceptable wave function? Asthe potential term is discontinuous in x, dψ(x)/dx need not be a continuous function. However,ψ(x) must be a continuous function, i.e. ψ(x = 0) = 0 and ψ(x = L) = 0. Taking the first conditionwe see that

ψ(x = 0) = A sin(0) +B cos(0) = 0 ⇒ B = 0. (58)

At x = L we haveψ(x = L) = A sin(kL) = 0 ⇒ kL = nπ. (59)


The wave function is therefore

ψ(x) = A sin(nπx

L

)forn = 1, 2, 3, . . . and 0 < x < L. (60)

The important point to note here is that the range of permitted wavefunctions is quantised. Wenext have to normalise the wave function over all space, i.e.∫ ∞

−∞ψ∗ψdx = 1

∫ 0

−∞ψ∗ψdx+

∫ L

0ψ∗ψdx+

∫ ∞

Lψ∗ψdx = 1

0 +∫ L

0A2 sin2

(nπx

L

)dx+ 0 = 1 (61)

We solve this integral by applying the change of variable u = nπx/L and writing∫ L

0A2 sin2

(nπx

L

)dx =

∫ nπ

0A2 sin2 u du

dx

du. (62)

The integral in Equation 61 then reduces to

A2 L

nπ

[−1

2cosu sinu+

u

2

]nπ

0= 1

A2 L

nπ

[−1

2cos(nπ) sin(nπ) +

nπ

2

]= 1

A =

√2

L. (63)

Hence the normalised wave function may be written

ψ(x) =

√2

Lsin

(nπx

L

)forn = 1, 2, 3, . . . and 0 < x < L. (64)

The wave function is exactly the same as a classical time independent standing wave and the particlein a box therefore corresponds to a stationary state.

T–Rex Example 6.7: Show that the wave function Ψn(x, t) for a particle in an infinitesquare well corresponds to a standing wave in a box.

To generate the time dependent wave function we form the quantity

Ψn(x, t) = ψ(x)eiωnt

=

√2

Lsin(knx)e

iωnt. (65)


However, sin(knx) can be written as

sin(knx) =eiknx − e−iknx

2i, (66)

so that the wave function becomes

Ψn(x, t) =

√2

L

(ei(knx−ωnt) − e−i(knx+ωnt)

2i

). (67)

This is simply the superposition of two waves, one travelling in the positive x–direction and onetravelling in the negative direction. A standing wave will result with an angular frequency ωn.

End of Example 6.7

The energy associated with each stationary state comes from a consideration of the wave number

kn =nπ

L=

√2mEn

h2 n = 1, 2, 3, . . . , (68)

where the subscript n indicates that kn and En depend upon the value of n – the principalquantum number. We may write the energy of each quantum state as

En = n2 π2h2

2mL2= n2 π2(hc)2

2(mc2)L2=

(eV nm)2

eV nm2= eV. (69)

Once again, we note that the possible energy values of the particle are quantised. In this case,energy quantisation arises solely from the boundary conditions. Each state available to the wavefunction ψn(x), is characterised by a unique energy state En, and probability density |ψn|2 (seeT-Rex Figure 6.3). A number of additional points should be emphasized

• The particle cannot have E = 0 (or at least only in the case where the particle does notexist but this is not an interesting possibility). The lowest available energy state is E1 cor-responding to n = 1. This is referred to as the ground state. Though this effect is onlynoticeable for quantum situations, the same rules holds in the macroscopic world. A tennisball constrained to lie within a court cannot have E = 0. However, in this case, E1 is so smallas to be unmeasureable in most practical circumstances. Once again, this view agrees withthe principle that macroscopic objects possess very large quantum numbers.

• In general we see that the energy of the confined particle is inversely proportional to both themass m and the square of the confinement scale L. Therefore, confining a given particle toa smaller region imposes a greater minimum energy. Taking the case of an electron we canwrite

E1 =π2(hc)2

2(mc2)L2=

(π2)(197.3 eV nm)2

(2)(511, 000 eV)(L2)≈ 0.4

(L/nm)2eV. (70)

Therefore, if we confine an electron in a one dimensional box of length 0.1 nm we obtain aground state energy E1 ≈ 40 eV, a reasonable approximation to atomic electron energies.


• The probability density P (x) to observe the particle at a given x is

P (x) = ψ∗ψ =2

Lsin2

(nπx

L

). (71)

In the classical limit this probability is P (x) = 1/L for 0 < x < L. For large values of n therewill be many wave oscillations within the box. As the average value of sin2 θ over one cycle is1/2. Therefore, in the limit of large quantum numbers we obtain the classical result.

T–Rex Example 6.8: Calculate the expected values of x, x2, p and p2 for a particle inan infinite square well existing in the first excited state.

The first excited state corresponds to n = 2. The wave funtion for this case is

ψ2(x) =

√2

Lsin

(2πx

L

)(72)

1. The expectation value of 〈x〉2 is

〈x〉2 =2

L

∫ L

0x sin2

(2πx

L

)dx. (73)

Once again, we solve this using a change of variable u = 2πx/L and noting that dx/du = L/2π.The intergral then becomes

〈x〉2 =2

L

(L

2π

)2 ∫ 2π

0u sin2 u du

=2

L

(L

2π

)2[u2

4− u

sin 2u

4− cos 2u

8

]2π

0

=2

L

(L

2π

)2 [π2 − 1

8+

1

8

]

=L

2. (74)

Note that the expected position of the particle is in the middle of the box although the valueof the wave function is zero at this point.

2. The expectation value of 〈x2〉2 is

〈x2〉2 =2

L

∫ L

0x2 sin2

(2πx

L

)dx

=(

1

3− 1

8π2

)L2 = 0.32L2, (75)


Note that the integral is solved with a change of variable to convert the integrand to a standardform and is then worked in a straightforward (if lengthy) manner.

3. The expectation of the momentum 〈p〉2 is given by

〈p〉2 = (−ih) 2

L

∫ L

0sin

(2πx

L

) [d

dxsin

(2πx

L

)]dx

= −4πih

L

∫ L

0sin

(2πx

L

)cos

(2πx

L

)dx

= 0. (76)

The average, or expected value of the momentum is zero – as the particle is moving left asoften as it is moving right.

4. However, the expectation value of 〈p2〉2 behaves in a different manner and is given by

〈p2〉2 =2

L

∫ L

0sin

(2πx

L

)(−ih d

dx

)(−ih d

dx

)sin

(2πx

L

)dx

= (−ih)2 2

L

∫ L

0sin

(2πx

L

)(2π

L

d

dx

)cos

(2πx

L

)dx

= −(−h2)8π2

L3

∫ L

0sin

(2πx

L

)sin

(2πx

L

)dx

=4π2h2

L2. (77)

This value can be compared to the value of E2, i.e.

E2 =4π2h2

2mL2=〈p2〉22m

, (78)

which is simply the statement E = p2/2m+ V , for which in this case we have V = 0.

Finally, what happens if we consider the uncertainty relation for this wave function? We can form

the uncertainty relation by noting that ∆x = (〈x2〉 − 〈x〉2)1/2and ∆p = (〈p2〉 − 〈p〉2)1/2

. In thiscase, we find that

∆p∆x =2πh

L

(L2

3− L2

4

)1/2

=2πh

L

L

2√

3= h

π√3>h

2. (79)


4.7 The finite square well potential

The infinite square well provides a useful first example of how to manipulate the time independentSchrodinger equation. However, infinite potentials are not very realistic. A more reasonable exampleof the potentials encountered in nature is the finite square well potential where we observe thefollowing

= V0 x ≤ 0 region I

V (x) = 0 0 < x < L region II

= V0 x ≥ L region III

We consider a particle of energy E < V0. In classical physics the particle is completely bound withinthe potential. However, within quantum theory we will see that there exists a possibility that theparticle can exist outside the potential well.

We begin by considering the time independent Schrodinger equation for regions I and III outsidethe square well

− h2

2m

d2ψ

dx2= (E − V0)ψ. (80)

We can re–write this equation using α2 = 2m(V0 − E)/h2, as positive constant, as

d2ψ

dx2= α2ψ. (81)

Solutions to this equation take the form eαx and e−αx, i.e. exponential terms rather than sinusoidal.We can use the boundary condition ψ(x) → 0 as x → ±∞ to select physically reasonable wavefunctions. In the region x > L we can reject the positive exponential term as otherwise we wouldhave ψ(x) → ∞ as x → ±∞. Similarly we can reject the negative exponential term at x < 0.Therefore we can write

ψI(x) = Aeαx x < 0

ψIII(x) = Be−αx x > L.

The constants A and B represent probability amplitudes that we must solve for using the boundaryconditions.

Within the potential well, in region II, the time independent Schrodinger equation takes the form

d2ψ

dx2= −k2ψ, (82)

where k =√

(2mE)/h2. The general sinusoidal wave function that satisfies this equation can bewritten as

ψII(x) = Ceikx +De−ikx 0 < x < L. (83)


Note that when we looked at the infinite square well, we were able to use the boundary conditionsto simplify this oscillatory solution to a sine term. However, with the finite square well, we nolonger have ψ(x) = for x = 0 or x = L.

We now use the boundary conditions to solve for the probability amplitudes A,B,C and D. We willnot solve for them in detail but rather illustrate the overall method. We require the wave functionψ(x) and its derivative dψ/dx be continuous functions of x. There are two “boundaries” in the finitesquare well (between regions I and II and between II and III). Therefore we can generate a totalof four equations to solve for four unknown probability amplitudes. We examine each condition inturn

• ψ(x) must be continuous. Therefore

ψI(x = 0) = ψII(x = 0)

A = C +D, (84)

and

ψII(x = L) = ψIII(x = L)

CeikL +De−ikL = Be−αL. (85)

• ψ′(x) = dψ/dx must be continuous. Therefore

ψ′I(x = 0) = ψ′II(x = 0)

αAeαx|x=0 = ikCikx|x=0 − ikDe−ikx|x=0

αA = ikC − ikD, (86)

and

ψ′II(x = L) = ψ′III(x = L)

ikCikL − ikDe−ikL = αBe−αL. (87)

We will not work further through the maths here. Instead, we emphasize a number of qualitativepoints.


• The form of the total wave function (I+II+III) for the finite square well is shown in T–RexFigure 6.5. The most important result is that where there exists a finite potential, thereexists a finite possibility to observe the particle outside the well. The greater the energy ofthe particle, the greater the probability of existing outside the potential well (see below). Thisis in contrast to the results derived for the infinite square well and for any classical analysisof a particle trapped in a box.

• Once again we note that the energy levels are quantized and that E = 0 is not permitted.

• If we compare wave solutions for the finite and infinite square wells, we see that the De Brogliewavelength of a given solution (n = 1, 2, 3, . . .) is longer in the finite well compared to theinfinite well. This in turn implies that the momenta and energy of each state are smaller forthe finite potential. In addition, the number of energy levels is curtailed to E < V0.

• The finite probability for the particle to exist outside the well is linked to the idea of quantumtunneling (more later). The extent to which the particle will be observed outside the well isgiven by the probability P (x)dx = ψ∗(x)ψ(x)dx. In regions III we can write

ψ2(x) = B2e−2αx ∝ e−2α = e−2√

2m(V0−E)h2

. (88)

With a similar expression for region I. We see that the probability to observe the particleoutside the well is smaller for larger potentials and also that the probability is larger forhigher energy states. The interval of distance δx over which the probability drops by a factore is equal to 1/2α, i.e.

δx =1

2α=

h

2√

2m(V0 − E). (89)

We call δx the penetration depth. In general, the presence of h in the numerator ensures thatδx corresponds to a very small distance.

Worked problem: consider a particle with an energy E bound within a finite square well ofheight V0 and width 2L such that −L ≤ x ≤ +L. As the potential energy is symmetric about themid-point of the well, the permitted stationary states will be either symmetric or anti-symmetricabout the mid-point.

1. Show that for E < V0 the boundary conditions restrict the permitted energies of symmetricwaves to be

k tan kL = α, (90)

where α =√

(2m/h2)(V0 − E), and k =√

2mE/h2 is the wave number of the oscillationinterior to the well.


Solution: The time independent Schrodinger equation reduces to

d2ψ

dx2=

2m

h2 (V (x)− E)ψ. (91)

Within the potential V (x) = 0 and we obtain solutions of the form sin kx and cos kx wherek2 = 2mE/h2. Waves that are symmetric about the mid-point of the well take the form

ψ(x) = A cos kx , (92)

for −L ≤ x ≤ +L. Outside the well, we have V (x) = V0 and solutions to the wave equationtake the form exp(±αx). Boundary conditions force us to reject growing exponentials. Inaddition, the condition that the total wave function be symmetric about the mid-point of thewell restricts us to

ψ(x) = Ce−α|x|, (93)

where α = 2m(V0−E)/h2 and for x < −L and x > L. We next apply the boundary conditionsat x = L. Continuity of ψ provides

A cos kL = Ce−αL. (94)

Continuity of dψ/dx provides

−Ak sin kL = −Cαe−αL. (95)

Dividing these two expressions leads to the desired expression for the permitted energies

k tan kL = α. (96)

2. Show that the energy condition above can be written as

k sec kL =

√2mV0

h. (97)

Solution: Inspection of k and α indicates that

k2 + α2 =2mV0

h2 (98)

We can therefore re–write the answer to part (1) as

k tan kL =

√2mV0

h2 − k2. (99)

We multiply this expression by L and square both sides. We note that tan2 θ+1 = sec2 θ andtake the square root of the resulting expression to obtain

k sec kL =

√2mV0

h, (100)

as required.


3. Apply this result to compute the ground state energy of an electron trapped with a defect ina crystal that displays an effective potential of 5eV and a width of 0.2nm.

Solution: Multiplying each side of the answer to part (2) by L generates

kL sec kL =

√2mV0L

h. (101)

We calculate the quantity

2mV0L

h2 =2mc2V0L

h2c2= (2)(511 keV)(5 eV)(0.1 nm)/(197.3 eV nm)2 = 1.3127, (102)

and note that we now have an equation of the form

kL sec kL =√

1.3127 = 1.1457. (103)

Expressions of this form can be solved numerically. A simple trial and error approach yieldskL = 0.799 from which we deduce that k = 7.99 nm−1. The energy of a non–relativisticelectron possessing thie wave number is

E =h2k2

2m=

(hc)2k2

2mc2=

(197 eV nm)2(7.99 nm−1)2

(2)(511 keV)= 2.432 eV. (104)

4.8 The three dimensional infinite square well

We are steadily improving our understanding of the Schrodinger equation to the point where wecan tackle realistic atomic physics problems. The next stage is to understand the application of theSchrodinger equation in three dimensions. We return to the three dimensional infinite square wellas a simple case.

Assuming that there will exist stationary states in three dimensions, we consider the time indepen-dent Schrodinger equation . However, the spatial component of the wave function is a function ofthree spatial coordinates, i.e. ψ = ψ(x, y, z). Applying the conservation of energy to the time inde-pendent Schrodinger equation and using our knowledge of operators, we can arrive at an expressionof for the Schrodinger equation in three dimensions. Starting with the conservation of energy wewrite

E = K + V =p2

2m+ V. (105)

Multiplying by the wave function ψ we obtain

Eψ =p2

2mψ + V ψ. (106)


To write p2 as an operator, we note that p2 = p2x + p2

y + p2z and therefore apply the momentum

operator in three dimensions. i.e.

px = −ih∂ψ∂x

px2 = −h2∂

2ψ

∂x2, (107)

with similar expressions for y and z. Therefore,

p2ψ = −h2

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)ψ = −h2∇2ψ, (108)

where

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2, (109)

is the Laplacian operator. Therefore, we may re–write our conservation of energy equation as

Eψ = − h2

2m∇2ψ + V ψ, (110)

which corresponds to the time independent Schrodinger equation in three dimensions.

Turning to the three dimensional infinite square well, we consider a box of sides equal to L1, L2, L3.The sides of the box are infinitely rigid such that V = 0 inside the box and V = ∞ outside. Theadditional point that guides us to a valid wave function is the fact that the box is free to move inany dimension within the box – the x, y, z components of the wave function must be independentof each other. Within the box the Schrodinger equation takes the form

− h2

2m∇2ψ = Eψ, (111)

and we guess a solution of the form

ψ(x, y, z) = A sin(k1x) sin(k2y) sin(k3z), (112)

where A is a normalisation constant and the wave numbers k1, k2, k3 are determined by applyingthe boundary conditions of the infinite well. In each dimension we have can apply the boundarycondition of the form ψ(x) = 0 for x = 0 and x = L1 such that

k1L1 = n1π ⇒ k1 =n1π

L1

, (113)


with similar expressions in the y and z dimensions. We calculate the permitted energy levels directlyfrom the Schrodinger equation . To do this we must first compute the spatial derivatives of thewave function (we show this explicitly for x only), i.e.

∂ψ

∂x=

∂

∂x[A sin(k1x) sin(k2y) sin(k3z)]

= k1A cos(k1x) sin(k2y) sin(k3z)

∂2ψ

∂x2=

∂

∂x[k1A cos(k1x) sin(k2y) sin(k3z)]

= −k21A sin(k1x) sin(k2y) sin(k3z)

= −k21ψ. (114)

As the derivatives for y and z are similar, the Schrodinger equation within the box becomes

h2

2m(k2

1 + k22 + k2

3)ψ = Eψ, (115)

which indicates that

E =h2

2m(k2

1 + k22 + k2

3). (116)

Inserting the values of k1, k2, k3 derived from the boundary conditions, we obtain the followingexpression for the permitted energy levels

En1n2n3 =π2h2

2m

(n2

1

L21

+n2

2

L22

+n2

3

L23

). (117)

Unsurprisingly, we now require three quantum numbers to describe the possible energy states of athree dimensional system.

In the case where we have a cubic well, i.e. L = L1 = L2 = L3, the energy states can be written as

En1n2n3 =π2h2

2mL2(n2

1 + n22 + n2

3). (118)

The ground state energy occurs when n1 = n2 = n3 = 1,

Egs =3π2h2

2mL2. (119)


There is only one combination of n1, n2, n3 that generates the ground state. What about the firstexcited state? The first excited state can be achieved with the following configurations

n1 n2 n3

1 1 21 2 12 1 1

(120)

Therefore, three different quantum states generate the same energy level – they are said to bedegenerate states. In this case the three dimensional symmetry of the potential box creates thedegeneracy. The degeneracy can be broken if the symmetry of the system is broken. Symmetrycan be broken in a number of ways: if the lengths of the side of the potential are unequal thenthe degenerate energy states will take different values. Symmetry can also be broken by externalfactors – an electric or a magnetic field say. As the electric /magnetic field is a vector, it introducesa preferred axis into the system such that wave terms aligned with the field will respond differentlyto wave terms aligned perpendicular. This is exactly what happens to the energy levels of an atomwhen subjected to an electric (Stark effect) or magnetic (Zeeman effect) field – the addition of avector force/potential breaks the degeneracy between the energy states.


Worked problem: Consider an electron trapped in a crystal defect modelled as a three dimensionalinfinite square well of length L = L1 = L2 = L3 = 0.5 nm.

1. Compute the probability of observing the ground state electron within x ± L/4, y ± L/4,z ± L/4.

Solution: To compute any probability based upon the wave function, we must first determinethe value of the normalisation constant, i.e.∫ L

0

∫ L

0

∫ L

0ψ∗(x, y, z)ψ(x, y, z) dx dy dz = 1. (121)

As the wave function involves no cross terms of the form sin(xy) etc., we can seperate the threedimensional wave function into three one dimensional functions, i.e. ψ(x, y, z) = f(x)g(y)h(z).This permits us to write the above normalisation expression into a more convenient form

A2

[∫ L

0sin2 k1x dx

] [∫ L

0sin2 k2y dy

] [∫ L

0sin2 k3z dz

]= 1. (122)

This equation is very similar to the case where we normalised the wave function for a one di-mensional infinite square well. In this case, each integral contained within the square bracketscontributes (L/2) such that

A2(L

2

)(L

2

)(L

2

)= 1 (123)

and

A =(

2

L

)3/2

, (124)

which is simply the cube of the normalisation constant determined for the one dimensionalcase. The average position of the particle in each dimension is simply L/2, once again, thesame result as for the one dimensional case. To compute the probability of observing theelectron within x± L/4, y ± L/4, z ± L/4, we therefore form the quantity

P (x± L/4, y ± L/4, z ± L/4) =∫ 3L/4

L/4

∫ 3L/4

L/4

∫ 3L/4

L/4ψ∗(x, y, z)ψ(x, y, z) dx dy dz

=(

2

L

)3[∫ 3L/4

L/4sin2 k1x dx

][plus similar terms for y and z].

We note that for the ground state electron k1 = π/L and we concentrate upon the form ofthe x integral, i.e.

∫ 3L/4

L/4sin2 k1x dx =

L

π

[x

2− 1

4sin 2x

]3L/4

L/4


=L

π

([3π

8− 1

4sin

(3π

2

)]−[π

8− 1

4sin

(π

2

)])

=L

4. (125)

Therefore, the probability integral over each dimension contributes a term L/4 and we mayre–write the three dimensional probability integral as

P (x± L/4, y ± L/4, z ± L/4) =(

2

L

)3(L4

)3

=1

8. (126)

2. Compute the ground state energy of the electron.

Solution: The ground state of a three dimensional well is described by the quantum numbersn1 = n2 = n3 = 1. Inserting these values into the expression for the energy levels of a particlein a three dimensional infinite square well, we obtain

E111 =3π2h2

2mL2=

3π2(hc)2

2(mc2)L2=

3π2

2

(197.3 eV nm)2

(511 keV)(0.5 nm)2 = 4.51 eV. (127)

3. What wavelength of photon will cause the ground state electron to make a transition to thefirst excited state?

Solution: The first excited state corresponds to an energy

E211 =π2h2

2mL2[22 + 12 + 12] = 2E111. (128)

Note that any one of the three possible first excited states would give the same answer. Inorder to cause the electron to change from the ground state to the first excited state thephoton energy must be

Ephoton =hc

λ= E211 − E111 = E111, (129)

Therefore,

λ =hc

E111

=1240 eV nm

×4.51 eV= 273 nm. (130)

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